What is a tight lower bound to sum^n_(i=1)(1)/(a+x_i) under the restrictions sum^n_(i=1)x_i=0 and sum^n_(i=1)x^2_i=a^2 ?

Mehlqv 2022-08-16 Answered
What is a tight lower bound to i = 1 n 1 a + x i under the restrictions i = 1 n x i = 0 and i = 1 n x i 2 = a 2 ?
Conjecture: due to the steeper rise of 1 a + x for negative x, one may keep those values as small as possible. So take n 1 values x i = q and x n = ( n 1 ) q to compensate for the first condition. The second one then gives a 2 = i = 1 n x i 2 = q 2 ( ( n 1 ) 2 + n 1 ) = q 2 n ( n 1 ). Hence,
i = 1 n 1 a + x i n 1 a ( 1 1 / n ( n 1 ) ) + 1 a ( 1 + ( n 1 ) / n ( n 1 ) )
should be the tight lower bound.
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Answers (1)

Jamir Young
Answered 2022-08-17 Author has 11 answers
Yes, this is indeed the minimum value (assuming a > 0).

Denote K = { x R n k = 1 n x k = 0 , k = 1 n x k 2 = a 2 } and let f ( x ) = k = 1 n ( a + x k ) 1 attain its minimum at x = x ¯ K (it does so, as a continuous function on { x K f ( x ) f ( 0 ) } which is compact). Then, by Lagrange multiplier theorem, there are λ 1 , λ 2 R such that for each k we have ( a + x ¯ k ) 2 = λ 1 + λ 2 x ¯ k . Then the positive numbers y k = a + x ¯ k are solutions of
y 2 ( λ 1 λ 2 a + λ 2 y ) = 1.
But this equation has at most two positive solutions. Thus, at most two values among x ¯ k are distinct, and in fact exactly two. So, let m values of x ¯ k equal b > 0, where 0 < m < n, and the remaining n m values equal c < 0. We get a system for b and c, solve it, and finally obtain
a f ( x ¯ ) = n + ( 1 1 n + n 2 m n m ( n m ) ) 1 .
The least possible value of this is at m = 1.

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